How to Convert Python List of Interger to List of Hex, Getting List Error "Struct.Error: Required Argument Is Not an Integer"

struct.pack raises "struct.error: required argument is not an integer"

ip_header = struct.pack('!BBHHHBBH4s4s' ,version_IHL, TOS, totalLength, ID,flags, TTL,protocolNR, checksum,sourceAddress,destinationAddress)

Both the B and H formats require integer arguments (or non-integer objects that implement the __index__ method) (see Format Characters).

The checksum argument is now of type bytes because you set it here before packing:

checksum = bytes(str(checksum),"utf-8")

And bytes objects do not implement the __index__ method.

You can check this using dir(checksum).

That's why you're getting the struct.error exception.

When packing a value x using one of the integer formats ('b', 'B',
'h', 'H', 'i', 'I', 'l', 'L', 'q', 'Q'), if x is outside the valid
range for that format then struct.error is raised.

Either:

• Use a different variable for the bytes(str(checksum),"utf-8") output
• Pass an int type object for the checksum value

Using big endian in struct pack Python with a Pointer

You may need to do this in two stages:

u1=1
u2=2
u3=3
u4=4
u5=5
point=6
data=7

# construct two separate buffers
b1 = struct.pack('>5I', u1, u2, u3, u4, u5)
b2 = struct.pack('P', point)

Then convert the pointer back to a regular uint, and then back to a pointer:

p2 = struct.unpack('I', b2)
b2a = struct.pack('>I', p2[0])
b1 = b1 + b2a
print(b1)

Output:

b'\x00\x00\x00\x01\x00\x00\x00\x02\x00\x00\x00\x03\x00\x00\x00\x04\x00\x00\x00\x05\x00\x00\x00\x06'

Hexlify an integer in Python

You can use str.format with x type:

>>> '{:x}'.format(123)
'7b'
>>> '{:08x}'.format(123)
'0000007b'

or using printf-style formatting:

>>> '%x' % 123
'7b'
>>> '%08x' % 123
'0000007b'

---

If you want to use binascii.hexlify, convert the int object to bytes using struct.pack:

>>> import struct
>>> import binascii
>>> struct.pack('i', 123)  # Replace 123 with `file_1_size_bytes`
b'{\x00\x00\x00'
>>> binascii.hexlify(struct.pack('i', 123))
b'7b000000'

You can control byte order using >, <, .. format specifier:

>>> binascii.hexlify(struct.pack('>i', 123))
b'0000007b'
>>> binascii.hexlify(struct.pack('<i', 123))
b'7b000000'

See Format characters - python struct module documentation

Python BCD splitting nibbles and concatenation 2 ints

you are interpreting the values as chars. Feeding chars to int() won't work, you have to feed the values as strings, like so: int("0x5E", 16). What you are attempting is in fact int(chr(int("0x5E", 16)),16), which is int("^",16) and will of course not work.

Do you expect these results?

makevalue('0x01', '0x5E') ->  350 0x15e 0b101011110
makevalue('0xFF', '0x00') ->  65280 0xff00 0b1111111100000000
makevalue('0x01', '0xFF') ->  511 0x1ff 0b111111111
makevalue('0xFF', '0xFF') ->  65535 0xffff 0b1111111111111111

If so, you can use this:

def makeValue(highbyte, lowbyte):
    return int(highbyte, 16)*256 + int(lowbyte, 16)

or the IMO more ugly and errorprone:

def makeValue(highbyte, lowbyte):
    return int(highbyte+lowbyte[2:], 16) # strips leading "0x" from lowbyte be4 concat

"OverflowError: Python int too large to convert to C long" on windows but not mac

You'll get that error once your numbers are greater than sys.maxsize:

>>> p = [sys.maxsize]
>>> preds[0] = p
>>> p = [sys.maxsize+1]
>>> preds[0] = p
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
OverflowError: Python int too large to convert to C long

You can confirm this by checking:

>>> import sys
>>> sys.maxsize
2147483647

To take numbers with larger precision, don't pass an int type which uses a bounded C integer behind the scenes. Use the default float:

>>> preds = np.zeros((1, 3))

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